An original Thinka practice paper modelled on the structure and difficulty of the Jun 2025 (V4) Cambridge International A Level Physics (9702) paper. Not affiliated with or reproduced from Cambridge.
卷一 選擇題
Forty multiple-choice questions. For each question there are four possible answers. Choose the one correct answer.
40 題目 · 40 分
題目 1 · 選擇題
1 分
What is the SI base unit of the intensity of a wave?
A.\(\text{kg m}^2 \text{s}^{-3}\)
B.\(\text{kg s}^{-3}\)
C.\(\text{kg m s}^{-2}\)
D.\(\text{kg m}^{-1} \text{s}^{-2}\)
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解題
Intensity is defined as power per unit area. Power is energy transferred per unit time, where energy (or work) is force multiplied by distance. In SI base units, force is \(\text{kg m s}^{-2}\), so energy is \(\text{kg m}^2 \text{s}^{-2}\), and power is \(\text{kg m}^2 \text{s}^{-3}\). Since intensity is power divided by area, its SI base unit is \(\frac{\text{kg m}^2 \text{s}^{-3}}{\text{m}^2} = \text{kg s}^{-3}\).
評分準則
1 mark for identifying the correct SI base unit as B.
題目 2 · 選擇題
1 分
A ball is thrown horizontally from the top of a cliff of height \(H\) with horizontal speed \(v\). It lands a horizontal distance \(D\) from the base of the cliff. Air resistance is negligible. If the ball is instead thrown with horizontal speed \(2v\) from a cliff of height \(2H\), what is the new horizontal distance it travels before landing?
A.\(\sqrt{2} D\)
B.\(2 D\)
C.\(2\sqrt{2} D\)
D.\(4 D\)
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解題
The time of flight \(t\) is determined by vertical motion: \(H = \frac{1}{2}gt^2 \implies t = \sqrt{\frac{2H}{g}}\). The horizontal distance is \(D = v t = v \sqrt{\frac{2H}{g}}\). For the second throw, the height is \(2H\), so the time of flight becomes \(t' = \sqrt{\frac{4H}{g}} = \sqrt{2}t\). With a horizontal speed of \(2v\), the new horizontal distance is \(D' = (2v) t' = 2v (\sqrt{2}t) = 2\sqrt{2} D\).
評分準則
1 mark for calculating the new horizontal distance as C.
題目 3 · 選擇題
1 分
A block of mass \(3m\) moving with speed \(u\) on a frictionless horizontal surface collides head-on with a stationary block of mass \(m\). The collision is perfectly elastic. What are the speeds of the two blocks after the collision?
A.speed of \(3m\) is \(0.25u\); speed of \(m\) is \(1.25u\)
B.speed of \(3m\) is \(0.50u\); speed of \(m\) is \(1.50u\)
C.speed of \(3m\) is \(0.75u\); speed of \(m\) is \(0.75u\)
D.speed of \(3m\) is \(0.50u\); speed of \(m\) is \(1.00u\)
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解題
In a perfectly elastic collision, the relative speed of approach equals the relative speed of separation: \(u - 0 = v_2 - v_1 \implies v_2 = u + v_1\), where \(v_1\) is the final speed of the mass \(3m\) and \(v_2\) is the final speed of the mass \(m\). By conservation of linear momentum: \(3m u = 3m v_1 + m v_2 \implies 3u = 3v_1 + (u + v_1) \implies 2u = 4v_1 \implies v_1 = 0.50u\). Substituting back gives \(v_2 = 1.50u\).
評分準則
1 mark for determining the correct final speeds as B.
題目 4 · 選擇題
1 分
A uniform cylindrical block of wood of length \(L\) and cross-sectional area \(A\) floats vertically in water of density \(\rho_w\) with a length \(h\) submerged. The block is then placed in an oil of density \(0.80\rho_w\). What is the submerged length of the block in the oil?
A.\(0.64 h\)
B.\(0.80 h\)
C.\(1.20 h\)
D.\(1.25 h\)
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解題
For a floating object, the upthrust is equal to the weight of the object. In water, Upthrust \(U_w = \rho_w A h g = W\). In oil, Upthrust \(U_{oil} = \rho_{oil} A h_{oil} g = W\). Setting the two equations equal: \(\rho_w A h g = 0.80\rho_w A h_{oil} g \implies h = 0.80 h_{oil} \implies h_{oil} = \frac{h}{0.80} = 1.25 h\).
評分準則
1 mark for calculating the correct submerged length in oil as D.
題目 5 · 選擇題
1 分
A water pump lifts \(20\text{ kg}\) of water per second through a vertical height of \(15\text{ m}\). The water is discharged from a pipe with a speed of \(8.0\text{ m s}^{-1}\). The electrical power input to the pump is \(4.5\text{ kW}\). What is the efficiency of the pump?
A.\(65\%\)
B.\(74\%\)
C.\(80\%\)
D.\(86\%\)
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解題
The useful power output consists of the rate of gaining gravitational potential energy and kinetic energy. \(\text{Power output} = \frac{m}{t} g h + \frac{1}{2} \frac{m}{t} v^2 = (20 \times 9.81 \times 15) + (0.5 \times 20 \times 8.0^2) = 2943 + 640 = 3583\text{ W}\). Efficiency \(= \frac{\text{Power output}}{\text{Power input}} = \frac{3583}{4500} \approx 0.796\), which is approximately \(80\%\).
評分準則
1 mark for calculating the efficiency as C.
題目 6 · 選擇題
1 分
Two wires \(P\) and \(Q\) are made of different materials. Wire \(P\) has diameter \(d\) and length \(L\). Wire \(Q\) has diameter \(2d\) and length \(2L\). Both wires are subjected to the same tensile force. The extension of wire \(P\) is twice that of wire \(Q\). What is the ratio of the Young modulus of \(P\) to the Young modulus of \(Q\)?
A.\(0.25\)
B.\(0.50\)
C.\(1.0\)
D.\(2.0\)
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解題
The Young modulus is given by \(E = \frac{F L}{A x}\). Since cross-sectional area \(A \propto d^2\), we can write \(E \propto \frac{L}{d^2 x}\). The ratio of the Young moduli is: \(\frac{E_P}{E_Q} = \frac{L_P}{L_Q} \times \frac{d_Q^2}{d_P^2} \times \frac{x_Q}{x_P} = \frac{1}{2} \times \frac{(2d)^2}{d^2} \times \frac{x_Q}{2x_Q} = \frac{1}{2} \times 4 \times \frac{1}{2} = 1.0\).
評分準則
1 mark for deriving the ratio of Young moduli as C.
題目 7 · 選擇題
1 分
A source of sound emitting a frequency of \(800\text{ Hz}\) moves at a constant speed towards a stationary observer. The frequency detected by the observer is \(860\text{ Hz}\). The speed of sound in air is \(340\text{ m s}^{-1}\). What is the speed of the source?
A.\(22\text{ m s}^{-1}\)
B.\(24\text{ m s}^{-1}\)
C.\(26\text{ m s}^{-1}\)
D.\(28\text{ m s}^{-1}\)
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解題
Using the Doppler effect formula for a source moving towards a stationary observer: \(f_o = f_s \left( \frac{v}{v - v_s} \right)\). Substituting the values: \(860 = 800 \left( \frac{340}{340 - v_s} \right) \implies 1.075 = \frac{340}{340 - v_s} \implies 340 - v_s = \frac{340}{1.075} \approx 316.3\text{ m s}^{-1}\). This gives \(v_s = 340 - 316.3 = 23.7\text{ m s}^{-1}\), which rounds to \(24\text{ m s}^{-1}\).
評分準則
1 mark for finding the source speed as B.
題目 8 · 選擇題
1 分
A copper wire of cross-sectional area \(A\) carries a current \(I\). The average drift speed of the free electrons is \(v\). A second copper wire has twice the diameter and carries a current of \(3I\). What is the average drift speed of the free electrons in the second wire?
A.\(0.375 v\)
B.\(0.75 v\)
C.\(1.5 v\)
D.\(3.0 v\)
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解題
The relationship between current and drift speed is given by \(I = n A v q\). Since the material is copper in both cases, the number density of free electrons \(n\) is constant. The cross-sectional area is proportional to the square of the diameter: \(A \propto d^2\). Thus, drift velocity is \(v \propto \frac{I}{d^2}\). For the second wire, \(v' \propto \frac{3I}{(2d)^2} = \frac{3}{4} \frac{I}{d^2}\), giving \(v' = 0.75 v\).
評分準則
1 mark for finding the correct drift velocity expression as B.
題目 9 · 選擇題
1 分
The rate of heat loss \(P\) from a cylindrical pipe is given by the equation:
\[P = k L \Delta T\]
where \(L\) is the length of the pipe, \(\Delta T\) is the temperature difference between the inside and outside of the pipe, and \(k\) is a constant.
What are the SI base units of the constant \(k\)?
A.\text{kg m s}^{-3} \text{K}^{-1}
B.\text{kg m}^2 \text{s}^{-3} \text{K}^{-1}
C.\text{kg s}^{-3} \text{K}^{-1}
D.\text{kg m s}^{-2} \text{K}^{-1}
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解題
To find the SI base units of the constant \(k\), we rearrange the equation to make \(k\) the subject:
\[k = \frac{P}{L \Delta T}\]
Now, we determine the SI base units for each quantity on the right-hand side:
- Power \(P\) is energy per unit time (\(\text{J s}^{-1}\)). In SI base units, \(1\text{ J} = 1\text{ kg m}^2 \text{s}^{-2}\), so: \[[P] = \text{kg m}^2 \text{s}^{-3}\] - Length \(L\) has the SI base unit: \[[L] = \text{m}\] - Temperature difference \(\Delta T\) has the SI base unit: \[[\Delta T] = \text{K}\]
Substituting these base units back into the equation for \(k\):
Award 1 mark for the correct answer A. Award 0 marks for incorrect options.
題目 10 · 選擇題
1 分
From a high balcony at height \(h\) above the ground, ball 1 is thrown vertically upwards with speed \(u\). At the same instant, ball 2 is thrown vertically downwards from the same balcony with the same speed \(u\). Air resistance is negligible and the acceleration of free fall is \(g\).
What is the difference in the time of flight between the two balls before they hit the ground?
A.\frac{u}{g}
B.\frac{2u}{g}
C.\sqrt{\frac{2h}{g}}
D.\frac{2u + \sqrt{2gh}}{g}
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解題
Let the vertically upward direction be positive, so acceleration \(a = -g\).
For ball 1 (thrown upwards): The initial velocity is \(+u\) and the displacement when it hits the ground is \(-h\). Using the equation of motion \(s = ut + \frac{1}{2}at^2\): \[-h = u t_1 - \frac{1}{2}g t_1^2 \implies \frac{1}{2}g t_1^2 - u t_1 - h = 0\]
For ball 2 (thrown downwards): The initial velocity is \(-u\) and the displacement is also \(-h\): \[-h = -u t_2 - \frac{1}{2}g t_2^2 \implies \frac{1}{2}g t_2^2 + u t_2 - h = 0\]
Subtracting the equation of ball 2 from that of ball 1: \[\left(\frac{1}{2}g t_1^2 - u t_1 - h\right) - \left(\frac{1}{2}g t_2^2 + u t_2 - h\right) = 0\] \[\frac{1}{2}g(t_1^2 - t_2^2) - u(t_1 + t_2) = 0\]
Factoring the difference of squares: \[\frac{1}{2}g(t_1 - t_2)(t_1 + t_2) - u(t_1 + t_2) = 0\]
Since \(t_1 + t_2 \neq 0\), we can divide the entire equation by \((t_1 + t_2)\): \[\frac{1}{2}g(t_1 - t_2) - u = 0\] \[t_1 - t_2 = \frac{2u}{g}\]
Thus, the difference in their times of flight is indeed \(\frac{2u}{g}\", which is independent of the balcony's height \)h\).
Therefore, the correct choice is B.
評分準則
Award 1 mark for the correct answer B. Award 0 marks for incorrect options.
題目 11 · 選擇題
1 分
A block of mass \(0.80\text{ kg}\) is initially sliding at a constant speed of \(3.0\text{ m s}^{-1}\) along a frictionless horizontal floor. A horizontal force \(F\) in the direction of motion is applied to the block. The force varies with time \(t\) as follows: - At \(t = 0\), \(F = 0\). - At \(t = 2.0\text{ s}\), \(F\) reaches a maximum of \(6.0\text{ N}\). - At \(t = 6.0\text{ s}\), \(F\) decreases linearly to \(0\).
What is the speed of the block at \(t = 6.0\text{ s}\)?
A.15.0\text{ m s}^{-1}
B.22.5\text{ m s}^{-1}
C.25.5\text{ m s}^{-1}
D.48.0\text{ m s}^{-1}
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解題
The impulse \(I\) acting on the block is given by the area under the force-time graph from \(t = 0\) to \(t = 6.0\text{ s}\).
Since the graph is a triangle with base \(6.0\text{ s}\) and peak height \(6.0\text{ N}\): \[I = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6.0\text{ s} \times 6.0\text{ N} = 18.0\text{ N s}\]
By the impulse-momentum theorem: \[I = \Delta p = m(v - u)\] where \(m = 0.80\text{ kg}\), \(u = 3.0\text{ m s}^{-1}\), and \(v\) is the final speed.
Substituting the values: \[18.0 = 0.80(v - 3.0)\] \[v - 3.0 = \frac{18.0}{0.80} = 22.5\text{ m s}^{-1}\] \[v = 22.5 + 3.0 = 25.5\text{ m s}^{-1}\]
Therefore, the correct choice is C.
評分準則
Award 1 mark for the correct answer C. Award 0 marks for incorrect options.
題目 12 · 選擇題
1 分
A U-tube of uniform cross-sectional area contains water of density \(\rho_w = 1000\text{ kg m}^{-3}\). An oil of density \(\rho_o = 800\text{ kg m}^{-3}\) is poured into the left arm of the U-tube until the oil column has a length of \(15.0\text{ cm}\). Both arms of the tube are open to the atmosphere.
What is the vertical difference in height between the top surface of the oil in the left arm and the top surface of the water in the right arm?
A.1.5\text{ cm}
B.3.0\text{ cm}
C.12.0\text{ cm}
D.15.0\text{ cm}
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解題
At the boundary interface between the oil and water in the left arm, the pressure must equal the pressure in the right arm at the same horizontal level.
Let \(h_w\) be the height of the water column in the right arm above this horizontal level, and \(h_o = 15.0\text{ cm}\) be the height of the oil column above this level.
The top surface of the oil is at height \(h_o = 15.0\text{ cm}\) above the reference interface. The top surface of the water is at height \(h_w = 12.0\text{ cm}\) above the reference interface.
The vertical difference in height \(\Delta h\) between the two surfaces is: \[\Delta h = h_o - h_w = 15.0\text{ cm} - 12.0\text{ cm} = 3.0\text{ cm}\]
Therefore, the correct choice is B.
評分準則
Award 1 mark for the correct answer B. Award 0 marks for incorrect options.
題目 13 · 選擇題
1 分
A motorized hoist is used to lift a crate of mass \(120\text{ kg}\) vertically upwards at a constant speed of \(1.5\text{ m s}^{-1}\). The efficiency of the hoist's motor system is \(65\%\).
What is the electrical power input to the motor? (Use \(g = 9.81\text{ m s}^{-2}\))
A.1.1\text{ kW}
B.1.8\text{ kW}
C.2.7\text{ kW}
D.4.2\text{ kW}
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解題
First, calculate the useful power output \(P_{\text{out}}\) required to lift the crate at constant speed: \[P_{\text{out}} = F v = m g v\] \[P_{\text{out}} = 120\text{ kg} \times 9.81\text{ m s}^{-2} \times 1.5\text{ m s}^{-1} = 1765.8\text{ W}\]
The efficiency \(\eta\) of the system is \(65\% = 0.65\).
The electrical power input \(P_{\text{in}}\) is given by: \[P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{1765.8\text{ W}}{0.65} \approx 2717\text{ W} = 2.7\text{ kW}\]
Therefore, the correct choice is C.
評分準則
Award 1 mark for the correct answer C. Award 0 marks for incorrect options.
題目 14 · 選擇題
1 分
Two wires X and Y are made of the same metal. Wire X has length \(L\) and diameter \(d\). Wire Y has length \(2L\) and diameter \(2d\). Both wires obey Hooke's law.
If the same tension \(F\) is applied to stretch both wires, what is the ratio of the elastic potential energy stored in X to the elastic potential energy stored in Y?
A.0.5
B.1.0
C.2.0
D.4.0
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解題
The elastic potential energy \(E\) stored in a wire under tension \(F\) is: \[E = \frac{1}{2} F x\] where \(x\) is the extension. Since both wires are subjected to the same tension \(F\), the ratio of the energy stored is: \[\frac{E_X}{E_Y} = \frac{x_X}{x_Y}\]
The extension \(x\) of a wire of length \(L\), cross-sectional area \(A\), and Young's modulus \(E_y\) is given by: \[x = \frac{F L}{A E_y}\] Since cross-sectional area is \(A = \frac{\pi d^2}{4}\), extension can be written as: \[x = \frac{4 F L}{\pi d^2 E_y} \propto \frac{L}{d^2}\]
Comparing wire X and wire Y: - For wire X: \(x_X \propto \frac{L}{d^2}\) - For wire Y: \(x_Y \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{L}{2d^2}\]
Thus, the ratio of extensions is: \[\frac{x_X}{x_Y} = \frac{L/d^2}{L/(2d^2)} = 2.0\]
Since the ratio of energy is equal to the ratio of extensions, \)\frac{E_X}{E_Y} = 2.0\).
Therefore, the correct choice is C.
評分準則
Award 1 mark for the correct answer C. Award 0 marks for incorrect options.
題目 15 · 選擇題
1 分
A progressive longitudinal wave is travelling from left to right through a medium. The displacement \(s\) of the particles in the medium is plotted against their equilibrium distance \(x\) from a reference point.
A positive value of \(s\) represents a displacement to the right.
Four points, P, Q, R, and S, are identified: - At P, \(s = 0\) and the displacement of particles immediately to the left is negative (shifted left) and immediately to the right is positive (shifted right). - At Q, the displacement \(s\) is at a positive maximum. - At R, \(s = 0\) and the displacement of particles immediately to the left is positive (shifted right) and immediately to the right is negative (shifted left). - At S, the displacement \(s\) is at a negative maximum.
Which point corresponds to a compression?
A.P
B.Q
C.R
D.S
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解題
A compression in a longitudinal wave is a region where the density and pressure of the medium are at their maximum. This occurs where nearby particles are displaced towards that region.
Let us analyze point R: - Particles immediately to the left of R have positive displacement, meaning they are shifted to the right (towards R). - Particles immediately to the right of R have negative displacement, meaning they are shifted to the left (towards R). - Since particles on both sides are shifted towards R, the particle concentration at R is maximized, representing a compression.
Let us analyze point P: - Particles immediately to the left of P have negative displacement (shifted left, away from P). - Particles immediately to the right of P have positive displacement (shifted right, away from P). - Since particles on both sides are shifted away from P, the particle concentration at P is minimized, representing a rarefaction.
Therefore, the correct choice is C.
評分準則
Award 1 mark for the correct answer C. Award 0 marks for incorrect options.
題目 16 · 選擇題
1 分
A battery of e.m.f. \(6.0\text{ V}\) and negligible internal resistance is connected in series with a fixed resistor of resistance \(200\ \Omega\) and a negative temperature coefficient (NTC) thermistor.
At an initial temperature of \(20\ ^\circ\text{C}\), the thermistor has a resistance of \(400\ \Omega\).
The temperature is then increased until the resistance of the thermistor decreases to \(100\ \Omega\).
What is the change in the potential difference across the fixed resistor?
A.A decrease of 1.0 V
B.A decrease of 2.0 V
C.An increase of 1.0 V
D.An increase of 2.0 V
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解題
Let's first calculate the initial potential difference across the fixed resistor \(V_{R,i}\):
The initial total resistance of the series circuit is: \[R_{\text{total, } i} = 200\ \Omega + 400\ \Omega = 600\ \Omega\]
The initial current \(I_i\) from the battery is: \[I_i = \frac{E}{R_{\text{total, } i}} = \frac{6.0\text{ V}}{600\ \Omega} = 0.010\text{ A}\]
The initial potential difference across the fixed resistor is: \[V_{R,i} = I_i \times R = 0.010\text{ A} \times 200\ \Omega = 2.0\text{ V}\]
Next, calculate the final potential difference across the fixed resistor \(V_{R,f}\) after the temperature increases:
The final total resistance of the circuit is: \[R_{\text{total, } f} = 200\ \Omega + 100\ \Omega = 300\ \Omega\]
The final current \(I_f\) in the circuit is: \[I_f = \frac{6.0\text{ V}}{300\ \Omega} = 0.020\text{ A}\]
The final potential difference across the fixed resistor is: \[V_{R,f} = I_f \times R = 0.020\text{ A} \times 200\ \Omega = 4.0\text{ V}\]
The change in potential difference across the fixed resistor is: \[\Delta V_R = V_{R,f} - V_{R,i} = 4.0\text{ V} - 2.0\text{ V} = +2.0\text{ V}\]
This is an increase of \(2.0\text{ V}\).
Therefore, the correct choice is D.
評分準則
Award 1 mark for the correct answer D. Award 0 marks for incorrect options.
題目 17 · 選擇題
1 分
What is the SI base unit of electrical resistivity?
The electrical resistance \(R\) is given by \(R = \frac{V}{I}\).
Since potential difference \(V\) is defined as work done per unit charge (\(V = \frac{W}{Q} = \frac{W}{I t}\)), we have: \[ R = \frac{W}{I^2 t} \]
In SI base units, work \(W\) (measured in joules) has units of \(\text{kg}\!\cdot\!\text{m}^2\!\cdot\!\text{s}^{-2}\). Therefore, the base units of \(R\) are: \[ [R] = \frac{\text{kg}\!\cdot\!\text{m}^2\!\cdot\!\text{s}^{-2}}{\text{A}^2\!\cdot\!\text{s}} = \text{kg}\!\cdot\!\text{m}^2\!\cdot\!\text{s}^{-3}\!\cdot\!\text{A}^{-2} \]
Resistivity \(\rho\) is related to resistance by \(\rho = \frac{R A}{l}\), where \(A\) is cross-sectional area and \(l\) is length. Thus: \[ [\rho] = [R] \times \frac{\text{m}^2}{\text{m}} = \text{kg}\!\cdot\!\text{m}^2\!\cdot\!\text{s}^{-3}\!\cdot\!\text{A}^{-2} \times \text{m} = \text{kg}\!\cdot\!\text{m}^3\!\cdot\!\text{s}^{-3}\!\cdot\!\text{A}^{-2} \]
評分準則
1 mark for the correct option. - Correctly identifies the base units of resistance: \(\text{kg}\!\cdot\!\text{m}^2\!\cdot\!\text{s}^{-3}\!\cdot\!\text{A}^{-2}\). - Correctly multiplies by \(\text{m}\) to get the base units of resistivity: \(\text{kg}\!\cdot\!\text{m}^3\!\cdot\!\text{s}^{-3}\!\cdot\!\text{A}^{-2}\).
題目 18 · 選擇題
1 分
A small stone is projected vertically upwards from the edge of a cliff of height \(H\) with an initial speed \(v\). The stone rises to its maximum height and then falls, eventually reaching the bottom of the cliff with speed \(2.5v\). Neglecting air resistance, what is the height \(H\) of the cliff?
A.\(\frac{3}{8}\frac{v^2}{g}\)
B.\(\frac{9}{8}\frac{v^2}{g}\)
C.\(\frac{21}{8}\frac{v^2}{g}\)
D.\(\frac{25}{8}\frac{v^2}{g}\)
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解題
Let us take the downward direction as positive. - The initial velocity of the stone is \(u = -v\). - The final velocity of the stone when it hits the bottom of the cliff is \(v_f = +2.5v\). - The acceleration is \(a = g\) (downwards). - The displacement of the stone is \(s = H\) (downwards).
Using the equations of motion: \[ v_f^2 = u^2 + 2as \] \[ (2.5v)^2 = (-v)^2 + 2gH \] \[ 6.25v^2 = v^2 + 2gH \] \[ 5.25v^2 = 2gH \] \[ H = \frac{5.25v^2}{2g} = \frac{21}{8}\frac{v^2}{g} \]
評分準則
1 mark for the correct option. - Correct application of \(v^2 = u^2 + 2as\) with correct signs for displacement and velocities. - Simplification of the ratio to obtain \(\frac{21}{8}\frac{v^2}{g}\).
題目 19 · 選擇題
1 分
A block of mass \(m\) moves with velocity \(u\) on a frictionless horizontal surface. It makes a direct head-on elastic collision with a stationary block of mass \(3m\). What are the velocities of the two blocks after the collision?
A.Mass \(m\) has velocity \(-0.5u\); mass \(3m\) has velocity \(+0.5u\)
B.Mass \(m\) has velocity \(-0.25u\); mass \(3m\) has velocity \(+0.75u\)
C.Mass \(m\) has velocity \(-0.5u\); mass \(3m\) has velocity \(+0.25u\)
D.Mass \(m\) has velocity \(-0.75u\); mass \(3m\) has velocity \(+0.25u\)
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解題
Let \(v_1\) be the velocity of block \(m\) and \(v_2\) be the velocity of block \(3m\) after the collision.
Since the collision is perfectly elastic, the relative speed of separation is equal to the relative speed of approach: \[ v_2 - v_1 = u - 0 \implies v_2 = v_1 + u \]
Using conservation of linear momentum: \[ m u = m v_1 + 3m v_2 \] \[ u = v_1 + 3(v_1 + u) \] \[ u = 4v_1 + 3u \] \[ -2u = 4v_1 \implies v_1 = -0.5u \]
Substitute \(v_1\) back to find \(v_2\): \[ v_2 = -0.5u + u = +0.5u \]
Thus, block \(m\) has velocity \(-0.5u\) and block \(3m\) has velocity \(+0.5u\).
評分準則
1 mark for the correct option. - Uses relative speed of approach = relative speed of separation. - Applies conservation of momentum and solves simultaneous equations.
題目 20 · 選擇題
1 分
A uniform horizontal shelf of length \(1.2\text{ m}\) and weight \(60\text{ N}\) is attached to a wall by a hinge. The other end is supported by a cable attached to a point on the shelf \(0.90\text{ m}\) from the wall. The cable makes an angle of \(30^\circ\) with the horizontal shelf. What is the tension in the cable?
A.\(40\text{ N}\)
B.\(60\text{ N}\)
C.\(80\text{ N}\)
D.\(120\text{ N}\)
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解題
Take moments about the hinge at the wall to keep the shelf in equilibrium. - The weight of the uniform shelf is \(60\text{ N}\) and acts at its midpoint, \(0.60\text{ m}\) from the hinge. - Clockwise moment = \(60\text{ N} \times 0.60\text{ m} = 36\text{ N}\!\cdot\!\text{ m}\).
- The tension \(T\) in the cable acts at a point \(0.90\text{ m}\) from the hinge at an angle of \(30^\circ\). - The vertical component of tension is \(T \sin(30^\circ)\). - Counter-clockwise moment = \(T \sin(30^\circ) \times 0.90\text{ m} = 0.45 T\).
Equating clockwise and counter-clockwise moments: \[ 36 = 0.45 T \implies T = \frac{36}{0.45} = 80\text{ N} \]
評分準則
1 mark for the correct option. - Identifies that the weight acts at the midpoint (0.60 m). - Correct moment equation: \(60 \times 0.60 = T \sin(30^\circ) \times 0.90\). - Resolves the tension correctly to find \(T = 80\text{ N}\).
題目 21 · 選擇題
1 分
A block of mass \(m\) is projected down a rough curved slope of length \(d\) from a vertical height \(h\). Its speed at the top of the slope is \(v\), and it just comes to rest at the bottom of the slope. What is the average resistive force \(F\) acting on the block as it slides down the slope?
A.\(F = \frac{m(gh - v^2)}{d}\)
B.\(F = \frac{m(2gh + v^2)}{2d}\)
C.\(F = \frac{m(gh + v^2)}{d}\)
D.\(F = \frac{m(2gh - v^2)}{2d}\)
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解題
The total mechanical energy at the start is the sum of gravitational potential energy and kinetic energy: \[ E_{\text{initial}} = mgh + \frac{1}{2}mv^2 \]
At the bottom of the slope, the block is at rest, so its final mechanical energy is zero: \[ E_{\text{final}} = 0 \]
The loss of mechanical energy is equal to the work done against the average resistive force \(F\) along the path of length \(d\): \[ W = F d = mgh + \frac{1}{2}mv^2 \]
Solving for \(F\): \[ F = \frac{mgh + \frac{1}{2}mv^2}{d} = \frac{m(2gh + v^2)}{2d} \]
評分準則
1 mark for the correct option. - Uses the principle of conservation of energy to write \(F d = \Delta E_{\text{k}} + \Delta E_{\text{p}}\). - Rearranges algebraically to find the correct expression.
題目 22 · 選擇題
1 分
Two wires, X and Y, are made of the same material. Wire X has length \(L\) and diameter \(d\). Wire Y has length \(2L\) and diameter \(2d\). Both wires are suspended vertically and subjected to the same tension \(F\). What is the ratio \(\frac{\text{extension of X}}{\text{extension of Y}}\)?
A.\(0.5\)
B.\(1.0\)
C.\(2.0\)
D.\(4.0\)
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解題
The Young modulus \(E\) of the material is given by: \[ E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{e / L} = \frac{F L}{A e} \]
Rearranging for extension \(e\): \[ e = \frac{F L}{A E} \]
Since the cross-sectional area \(A = \frac{\pi d^2}{4}\), the extension is proportional to: \[ e \propto \frac{L}{d^2} \]
Let the extension of wire X be \(e_{\text{X}}\) and wire Y be \(e_{\text{Y}}\): \[ \frac{e_{\text{X}}}{e_{\text{Y}}} = \frac{L_{\text{X}}}{L_{\text{Y}}} \times \left( \frac{d_{\text{Y}}}{d_{\text{X}}} \right)^2 \]
Substitute the given values: \[ \frac{e_{\text{X}}}{e_{\text{Y}}} = \frac{L}{2L} \times \left( \frac{2d}{d} \right)^2 = \frac{1}{2} \times 4 = 2.0 \]
評分準則
1 mark for the correct option. - Expresses extension in terms of length, area, Young modulus, and force. - Shows that extension is proportional to \(L/d^2\). - Calculates the ratio correctly as 2.0.
題目 23 · 選擇題
1 分
A train sounding a whistle of constant frequency \(800\text{ Hz}\) is traveling towards a stationary observer at a constant speed. The speed of sound in air is \(340\text{ m s}^{-1}\). The observer detects a frequency of \(880\text{ Hz}\). What is the speed of the train?
A.\(31\text{ m s}^{-1}\)
B.\(34\text{ m s}^{-1}\)
C.\(38\text{ m s}^{-1}\)
D.\(43\text{ m s}^{-1}\)
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解題
For a source moving towards a stationary observer, the Doppler-shifted frequency \(f_o\) is given by: \[ f_o = f_s \left( \frac{v}{v - v_s} \right) \]
Where: - \(f_o = 880\text{ Hz}\) (observed frequency) - \(f_s = 800\text{ Hz}\) (source frequency) - \(v = 340\text{ m s}^{-1}\) (speed of sound) - \(v_s\) is the speed of the source (train)
1 mark for the correct option. - Recalls and uses correct formula for Doppler effect with moving source. - Calculates the speed of the train correctly to two significant figures.
題目 24 · 選擇題
1 分
A potential divider circuit consists of a fixed resistor of resistance \(3000\ \Omega\) connected in series with a thermistor across a \(12\text{ V}\) power supply of negligible internal resistance. The output voltage \(V_{\text{out}}\) is measured across the thermistor. At temperature \(T_1\), the thermistor has a resistance of \(6000\ \Omega\). At a higher temperature \(T_2\), its resistance is \(1500\ \Omega\). What is the change in the output voltage \(V_{\text{out}}\) as the temperature rises from \(T_1\) to \(T_2\)?
A.It increases by \(2\text{ V}\)
B.It decreases by \(2\text{ V}\)
C.It increases by \(4\text{ V}\)
D.It decreases by \(4\text{ V}\)
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解題
Using the potential divider equation, the output voltage across the thermistor is: \[ V_{\text{out}} = V_{\text{in}} \left( \frac{R_{\text{th}}}{R + R_{\text{th}}} \right) \]
At temperature \(T_1\): \[ V_{\text{out1}} = 12 \left( \frac{6000}{3000 + 6000} \right) = 12 \times \frac{6000}{9000} = 8.0\text{ V} \]
At temperature \(T_2\): \[ V_{\text{out2}} = 12 \left( \frac{1500}{3000 + 1500} \right) = 12 \times \frac{1500}{4500} = 4.0\text{ V} \]
So the output voltage decreases by \(4\text{ V}\).
評分準則
1 mark for the correct option. - Calculates initial output voltage at \(T_1\) as \(8.0\text{ V}\). - Calculates final output voltage at \(T_2\) as \(4.0\text{ V}\). - Correctly determines that the voltage decreases by \(4.0\text{ V}\).
題目 25 · 選擇題
1 分
The volume rate of flow of a liquid of viscosity \( \eta \) through a capillary tube of length \( L \) and radius \( r \) under a pressure difference \( \Delta P \) is given by:
where \( C \) is a dimensionless constant. What is the value of the exponent \( y \)?
A.1
B.2
C.3
D.4
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解題
We use dimensional analysis to find the values of the exponents.
1. The SI base units of each quantity are: - Volume flow rate \( [V/t] = \text{m}^3\ \text{s}^{-1} \) - Pressure gradient \( [\Delta P / L] = \frac{\text{N}\ \text{m}^{-2}}{\text{m}} = \text{kg}\ \text{m}^{-2}\ \text{s}^{-2} \) - Radius \( [r] = \text{m} \) - Dynamic viscosity \( [\eta] = \text{kg}\ \text{m}^{-1}\ \text{s}^{-1} \)
2. Substitute these base units into the homogeneous equation: \[ \text{m}^3\ \text{s}^{-1} = (\text{kg}\ \text{m}^{-2}\ \text{s}^{-2})^x \cdot (\text{m})^y \cdot (\text{kg}\ \text{m}^{-1}\ \text{s}^{-1})^z \]
3. Equate the exponents on both sides of the equation: - For mass (\( \text{kg} \)): \( 0 = x + z \implies z = -x \) - For time (\( \text{s} \)): \( -1 = -2x - z \implies -1 = -2x - (-x) \implies x = 1 \), which means \( z = -1 \). - For length (\( \text{m} \)): \( 3 = -2x + y - z \implies 3 = -2(1) + y - (-1) \implies y = 4 \).
Therefore, the value of the exponent \( y \) is 4.
評分準則
1 mark for the correct option. - Award 1 mark for calculating the correct exponent through dimensional analysis. - Incorrect choices represent errors in deriving base units of pressure gradient or viscosity.
題目 26 · 選擇題
1 分
A radio-controlled model car travels along a straight line. The variation with time \( t \) of its velocity \( v \) is as follows: - From \( t = 0 \) to \( t = 2.0\text{ s} \), its velocity increases uniformly from \( 0 \) to \( 6.0\text{ m s}^{-1} \). - From \( t = 2.0\text{ s} \) to \( t = 5.0\text{ s} \), its velocity remains constant at \( 6.0\text{ m s}^{-1} \). - From \( t = 5.0\text{ s} \) to \( t = 9.0\text{ s} \), its velocity decreases at a constant rate from \( 6.0\text{ m s}^{-1} \) to \( -6.0\text{ m s}^{-1} \). - From \( t = 9.0\text{ s} \) to \( t = 12.0\text{ s} \), its velocity remains constant at \( -6.0\text{ m s}^{-1} \).
What is the average velocity of the car during the \( 12.0\text{ s} \) interval?
A.0.50 m s⁻¹
B.2.5 m s⁻¹
C.4.0 m s⁻¹
D.4.5 m s⁻¹
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解題
The average velocity is defined as the total displacement divided by the total time.
1. Calculate the positive displacement (from \( t = 0 \) to \( t = 7.0\text{ s} \), where \( v \ge 0 \)): - From \( t = 0 \) to \( t = 2.0\text{ s} \): \( s_1 = \frac{1}{2} \times 2.0 \times 6.0 = 6.0\text{ m} \) - From \( t = 2.0\text{ s} \) to \( t = 5.0\text{ s} \): \( s_2 = (5.0 - 2.0) \times 6.0 = 18.0\text{ m} \) - From \( t = 5.0\text{ s} \) to \( t = 7.0\text{ s} \) (when \( v \) decreases from \( 6.0\text{ m s}^{-1} \) to \( 0 \)): \( s_3 = \frac{1}{2} \times (7.0 - 5.0) \times 6.0 = 6.0\text{ m} \) Total positive displacement = \( 6.0 + 18.0 + 6.0 = 30.0\text{ m} \).
2. Calculate the negative displacement (from \( t = 7.0\text{ s} \ to \ t = 12.0\text{ s} \), where \( v \le 0 \)): - From \( t = 7.0\text{ s} \) to \( t = 9.0\text{ s} \) (when \( v \) decreases from \( 0 \) to \( -6.0\text{ m s}^{-1} \)): \( s_4 = \frac{1}{2} \times (9.0 - 7.0) \times (-6.0) = -6.0\text{ m} \) - From \( t = 9.0\text{ s} \) to \( t = 12.0\text{ s} \): \( s_5 = (12.0 - 9.0) \times (-6.0) = -18.0\text{ m} \) Total negative displacement = \( -6.0 - 18.0 = -24.0\text{ m} \).
3. Calculate the total displacement: \( s_{\text{total}} = 30.0 - 24.0 = 6.0\text{ m} \).
4. Calculate the average velocity: \( v_{\text{avg}} = \frac{s_{\text{total}}}{t_{\text{total}}} = \frac{6.0\text{ m}}{12.0\text{ s}} = 0.50\text{ m s}^{-1} \).
(Note: Option D, \( 4.5\text{ m s}^{-1} \), represents the average speed, which is a common error.)
評分準則
1 mark for the correct option. - Award 1 mark for correctly determining total displacement as the net area under the v-t graph and dividing by total time. - Deduct/distractor options correspond to finding average speed (D) or incorrect integration steps.
題目 27 · 選擇題
1 分
Two gliders, X and Y, are on a horizontal, frictionless air track. Glider X has mass \( 3m \) and moves to the right with speed \( u \). Glider Y has mass \( m \) and moves to the left with speed \( 2u \).
The gliders collide. After the collision, glider Y moves to the right with speed \( v \).
What is the magnitude of the change in momentum of glider X?
A.m(2u - v)
B.m(2u + v)
C.3m(u - v)
D.3m(2u + v)
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解題
By Newton's third law, the force exerted by glider X on Y is equal and opposite to the force exerted by Y on X. Consequently, the change in momentum of glider X (\( \Delta p_X \)) is equal in magnitude and opposite in direction to the change in momentum of glider Y (\( \Delta p_Y \)).
1. Let the rightward direction be positive. - Initial momentum of glider Y: \( p_{Y,i} = m(-2u) = -2mu \) - Final momentum of glider Y: \( p_{Y,f} = m(+v) = +mv \)
2. Calculate the change in momentum of glider Y: \( \Delta p_Y = p_{Y,f} - p_{Y,i} = mv - (-2mu) = m(2u + v) \)
3. Since \( |\Delta p_X| = |\Delta p_Y| \), the magnitude of the change in momentum of glider X is: \( |\Delta p_X| = m(2u + v) \).
評分準則
1 mark for the correct option. - Award 1 mark for applying the conservation of momentum / Newton's third law to relate the momentum change of X to that of Y. - Distractor options represent signs errors in momentum directions.
題目 28 · 選擇題
1 分
A uniform wooden block of mass \( M \) floats in water of density \( \rho_w \). A small metal object of mass \( m \) and volume \( V \) is suspended by a light string from the bottom of the block, so that the metal object is completely submerged in the water while the wooden block still floats partially submerged.
Compared to the block floating alone, what is the change in the volume of the wooden block that is submerged in the water?
A.An increase of \( \frac{m}{\rho_w} \)
B.An increase of \( \frac{m}{\rho_w} - V \)
C.An increase of \( \frac{m}{\rho_w} + V \)
D.An increase of \( \frac{M+m}{\rho_w} - V \)
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解題
1. When the wooden block floats alone: - Total downward force = \( M g \) - Upthrust \( U_1 = \rho_w V_{\text{sub},1} g \) - Since it floats, \( U_1 = M g \implies V_{\text{sub},1} = \frac{M}{\rho_w} \).
2. When the metal object is suspended below the block: - Total downward force = \( (M + m) g \) - Both the block (partially) and the metal object (fully) are submerged, so they both experience upthrust. - Total Upthrust \( U_2 = U_{\text{block}} + U_{\text{metal}} = \rho_w V_{\text{sub},2} g + \rho_w V g \) - Since the system floats, \( U_2 = (M + m)g \implies \rho_w V_{\text{sub},2} g + \rho_w V g = (M + m)g \) - Simplify to find \( V_{\text{sub},2} \): \( V_{\text{sub},2} = \frac{M + m}{\rho_w} - V \)
3. The change in the submerged volume of the wooden block is: \( \Delta V_{\text{sub}} = V_{\text{sub},2} - V_{\text{sub},1} = \left( \frac{M + m}{\rho_w} - V \right) - \frac{M}{\rho_w} = \frac{m}{\rho_w} - V \).
Since the density of the metal object is greater than that of water (\( m/V > \rho_w \)), the term \( \frac{m}{\rho_w} - V \) is positive, indicating an increase.
評分準則
1 mark for the correct option. - Award 1 mark for correctly setting up Archimedes' equations in both states and solving for the difference in block submerged volume. - Distractors miss the upthrust contribution acting directly on the metal object.
題目 29 · 選擇題
1 分
An electric pump is used to pump water from a well of depth \( 12\text{ m} \) and discharge it through a horizontal pipe of cross-sectional area \( 5.0 \times 10^{-3}\text{ m}^2 \) at a speed of \( 4.0\text{ m s}^{-1} \).
The density of water is \( 1000\text{ kg m}^{-3} \).
What is the minimum useful power of the pump? (Take \( g = 9.81\text{ m s}^{-2} \))
A.0.16 kW
B.2.2 kW
C.2.4 kW
D.2.5 kW
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解題
The pump must do work to lift the water (increasing its gravitational potential energy) and to accelerate the water (giving it kinetic energy).
1. Calculate the rate of mass flow of water \( \frac{dm}{dt} \): \( \frac{dm}{dt} = \rho A v = 1000\text{ kg m}^{-3} \times (5.0 \times 10^{-3}\text{ m}^2) \times 4.0\text{ m s}^{-1} = 20\text{ kg s}^{-1} \)
2. Calculate the rate of increase of gravitational potential energy (useful power for lifting): \( P_{\text{pe}} = \frac{dm}{dt} g h = 20\text{ kg s}^{-1} \times 9.81\text{ m s}^{-2} \times 12\text{ m} = 2354.4\text{ W} \)
3. Calculate the rate of increase of kinetic energy: \( P_{\text{ke}} = \frac{1}{2} \frac{dm}{dt} v^2 = \frac{1}{2} \times 20\text{ kg s}^{-1} \times (4.0\text{ m s}^{-1})^2 = 160\text{ W} \)
4. Calculate the total useful power output required: \( P_{\text{total}} = P_{\text{pe}} + P_{\text{ke}} = 2354.4\text{ W} + 160\text{ W} = 2514.4\text{ W} \approx 2.5\text{ kW} \).
評分準則
1 mark for the correct option. - Award 1 mark for evaluating both potential energy per second and kinetic energy per second and summing them. - Distractor C misses kinetic energy.
題目 30 · 選擇題
1 分
Two wires, X and Y, are made of the same metal. Wire X has length \( L \) and radius \( r \). Wire Y has length \( 2L \) and radius \( 2r \).
Both wires are suspended vertically and support loads. When wire X is loaded with a weight \( W \), the elastic energy stored in the wire is \( E \). Under this condition, both wires stretch within their elastic limits.
When wire Y is loaded with a weight \( 2W \), what is the elastic energy stored in wire Y?
A.½ E
B.E
C.2E
D.4E
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解題
1. Express the elastic energy stored in terms of force \( F \), length \( L \), cross-sectional area \( A \), and Young modulus \( E_Y \): \[ E_{\text{stored}} = \frac{1}{2} F e \] Since extension \( e = \frac{F L}{A E_Y} \), we have: \[ E_{\text{stored}} = \frac{F^2 L}{2 A E_Y} \]
2. For wire X: - Load \( F_X = W \), Length \( L_X = L \), Area \( A_X = \pi r^2 \) \[ E_X = \frac{W^2 L}{2 \pi r^2 E_Y} = E \]
1 mark for the correct option. - Award 1 mark for expressing strain energy in terms of Young modulus and geometry, then finding the correct algebraic ratio. - Distractor errors include incorrect scaling of radius squared or incorrect substitution.
題目 31 · 選擇題
1 分
A stationary observer detects sound from a siren on a vehicle moving directly towards them. The siren emits sound of a constant frequency.
The vehicle passes the observer and continues to move away from them at the same constant speed.
The ratio of the frequency of sound detected as the vehicle approaches to the frequency detected as it recedes is \( 1.125 \).
What is the speed of the vehicle? (The speed of sound in air is \( 340\text{ m s}^{-1} \).)
A.15 m s⁻¹
B.20 m s⁻¹
C.25 m s⁻¹
D.30 m s⁻¹
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解題
Using the Doppler effect formulas: 1. Observed frequency when approaching: \[ f_{\text{app}} = f \left( \frac{v}{v - v_s} \right) \] 2. Observed frequency when receding: \[ f_{\text{rec}} = f \left( \frac{v}{v + v_s} \right) \] where \( v = 340\text{ m s}^{-1} \) is the speed of sound, and \( v_s \) is the speed of the vehicle.
3. Set up the ratio: \[ \frac{f_{\text{app}}}{f_{\text{rec}}} = \frac{v + v_s}{v - v_s} = 1.125 = \frac{9}{8} \]
1 mark for the correct option. - Award 1 mark for establishing the Doppler frequency ratio and correctly solving for the source speed. - Distractor choices represent minor algebraic errors in solving the ratio equation.
題目 32 · 選擇題
1 分
A junction in a circuit has three wires connected to it. - Wire 1 carries a current of \( 3.0\text{ A} \) into the junction. - Wire 2 is connected to a resistor of resistance \( 4.0\ \Omega \) which has a potential difference of \( 8.0\text{ V} \) across it.
What are the possible values for the current in Wire 3?
A.1.0 A or 2.0 A
B.1.0 A or 5.0 A
C.2.0 A or 5.0 A
D.3.0 A or 5.0 A
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解題
1. By Ohm's law, the magnitude of the current in Wire 2 is: \[ I_2 = \frac{V}{R} = \frac{8.0\text{ V}}{4.0\ \Omega} = 2.0\text{ A} \] Since the direction of this current is not specified, it could flow either *into* or *out of* the junction.
2. Apply Kirchhoff's First Law (conservation of charge: total current entering = total current leaving): - **Case 1**: Current in Wire 2 flows *into* the junction. - Current entering = \( 3.0\text{ A} \text{ (Wire 1)} + 2.0\text{ A} \text{ (Wire 2)} = 5.0\text{ A} \). - Current leaving in Wire 3 must therefore be \( 5.0\text{ A} \).
Thus, the possible current values in Wire 3 are \( 1.0\text{ A} \) or \( 5.0\text{ A} \).
評分準則
1 mark for the correct option. - Award 1 mark for finding the current in Wire 2 (2.0 A) and considering both directional cases of Kirchhoff's First Law to find the potential outputs (1.0 A or 5.0 A). - Distractor choices contain incorrect single current calculations or misuse of current values.
題目 33 · 選擇題
1 分
The rate of heat transfer \(\frac{Q}{t}\) (where \(Q\) is energy and \(t\) is time) through a material of thermal conductivity \(k\), cross-sectional area \(A\), and temperature gradient \(\frac{\Delta T}{\Delta x}\) (where \(T\) is temperature and \(x\) is thickness) is given by the equation: \\ \\ \(\frac{Q}{t} = k A \frac{\Delta T}{\Delta x}\)\\ \\ What are the SI base units of thermal conductivity \(k\)?
First, find the SI base units of each quantity in the equation:\ - Rate of heat transfer \(\frac{Q}{t}\) is power, which has SI base units of \(\text{kg m}^2 \text{s}^{-3}\).\ - Cross-sectional area \(A\) has SI base units of \(\text{m}^2\).\ - Temperature gradient \(\frac{\Delta T}{\Delta x}\) has SI base units of \(\text{K m}^{-1}\).\ \ Rearranging the equation for \(k\):\ \(k = \frac{Q/t}{A (\Delta T / \Delta x)}\\\n\nSubstituting the base units into this rearranged equation:\n\)[k] = \frac{\text{kg m}^2 \text{s}^{-3}}{\text{m}^2 \cdot \text{K m}^{-1}} = \frac{\text{kg m}^2 \text{s}^{-3}}{\text{m K}} = \text{kg m s}^{-3} \text{K}^{-1}\)
評分準則
1 mark for the correct option A. Method: Identify base units of power, area, and temperature gradient, then solve for the base units of the constant.
題目 34 · 選擇題
1 分
A ball is thrown vertically upwards from a balcony of height \(h\) with an initial speed of \(12.0\text{ m s}^{-1}\). It strikes the ground below the balcony with a speed of \(24.0\text{ m s}^{-1}\). Air resistance is negligible. What is the height \(h\) of the balcony?
A.\(11.0\text{ m}\)
B.\(22.0\text{ m}\)
C.\(29.4\text{ m}\)
D.\(36.7\text{ m}\)
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解題
Using the equations of motion with constant acceleration \(a = g = 9.81\text{ m s}^{-2}\) downwards:\ Taking downwards as the positive direction:\ - Initial velocity \(u = -12.0\text{ m s}^{-1}\) (upwards)\ - Final velocity \(v = 24.0\text{ m s}^{-1}\) (downwards)\ - Displacement \(s = h\)\ \ Using the equation:\ \(v^2 = u^2 + 2as\)\ \((24.0)^2 = (-12.0)^2 + 2(9.81)h\)\ \(576 = 144 + 19.62 h\)\ \(432 = 19.62 h\)\ \(h \approx 22.0\text{ m}\)
評分準則
1 mark for the correct option B. Method: Use of \(v^2 = u^2 + 2as\) with correct signs for velocities to find displacement.
題目 35 · 選擇題
1 分
Sand falls vertically onto a horizontal conveyor belt at a constant rate of \(15.0\text{ kg s}^{-1}\). The belt is moving horizontally at a constant speed of \(2.4\text{ m s}^{-1}\). What is the horizontal force required to keep the belt moving at this constant speed, and at what rate is heat generated due to friction between the sand and the belt?
1. The horizontal force \(F\) required to accelerate the falling sand of mass rate \(\frac{\Delta m}{\Delta t}\) to the horizontal speed \(v\) of the belt is:\ \(F = v \frac{\Delta m}{\Delta t} = (2.4\text{ m s}^{-1}) \times (15.0\text{ kg s}^{-1}) = 36.0\text{ N}\\n\\n2. The total rate of doing work (power input) by the conveyor belt drive motor is:\n\)P_{\text{total}} = F v = 36.0\text{ N} \times 2.4\text{ m s}^{-1} = 86.4\text{ W}\\\n\\\n3. The rate at which the sand gains kinetic energy is:\\n\(P_{\text{KE}} = \frac{1}{2} \frac{\Delta m}{\Delta t} v^2 = \frac{1}{2} \times 15.0\text{ kg s}^{-1} \times (2.4\text{ m s}^{-1})^2 = 43.2\text{ W}\\n\\n4. The rate of heat generation due to friction is the difference between the total power supplied and the rate of gain of kinetic energy:\n\)P_{\text{heat}} = P_{\text{total}} - P_{\text{KE}} = 86.4\text{ W} - 43.2\text{ W} = 43.2\text{ W}\)
評分準則
1 mark for the correct option C. Method: Use rate of change of momentum to find force, then compute total power input and subtract the rate of kinetic energy increase to find the rate of heat generation.
題目 36 · 選擇題
1 分
A uniform plank of weight \(150\text{ N}\) and length \(3.0\text{ m}\) is supported by two pivots. Pivot A is at a distance of \(0.6\text{ m}\) from the left-hand end. Pivot B is at a distance of \(2.2\text{ m}\) from the left-hand end. A load \(W\) is placed at the extreme right-hand end of the plank. As \(W\) is gradually increased, the plank begins to tilt. At what value of \(W\) does the plank lift off pivot A?
A.\(75\text{ N}\)
B.\(131\text{ N}\)
C.\(225\text{ N}\)
D.\(300\text{ N}\)
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解題
When the plank is on the point of lifting off pivot A, the normal reaction force at pivot A is zero, and the plank is in equilibrium about pivot B.\ \ 1. Find the position of the centre of gravity (CoG):\ Since the plank is uniform, its weight of \(150\text{ N}\) acts at its midpoint, which is \(1.5\text{ m}\) from the left end.\ \ 2. Calculate the perpendicular distance of each force from the pivot B (located at \(2.2\text{ m}\) from the left end):\ - Distance of the weight from pivot B = \(2.2\text{ m} - 1.5\text{ m} = 0.7\text{ m}\) (to the left).\ - Distance of the load \(W\) (located at \(3.0\text{ m}\) from the left end) from pivot B = \(3.0\text{ m} - 2.2\text{ m} = 0.8\text{ m}\) (to the right).\ \ 3. Apply the principle of moments about pivot B:\ \(\text{Clockwise Moment} = \text{Anticlockwise Moment}\\n\)W \times 0.8\text{ m} = 150\text{ N} \times 0.7\text{ m}\\\n\(0.8 W = 105\text{ N m}\\n\)W = 131.25\text{ N} \approx 131\text{ N}\)
評分準則
1 mark for the correct option B. Method: Identify pivot B as the fulcrum, determine distances of center of gravity and the load from pivot B, and apply the principle of moments.
題目 37 · 選擇題
1 分
An electric pump is used to draw water from a well of depth \(12.0\text{ m}\) and eject it at the surface with a speed of \(8.0\text{ m s}^{-1}\). The pump delivers water at a rate of \(5.0\text{ kg s}^{-1}\). The overall efficiency of the pump system is \(75\%\). What is the electrical power input to the motor?
A.\(0.56\text{ kW}\)
B.\(0.75\text{ kW}\)
C.\(0.79\text{ kW}\)
D.\(1.0\text{ kW}\)
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解題
1. Find the useful power output of the pump per second:\ - Rate of gain of gravitational potential energy: \\ \(P_{\text{GPE}} = \frac{\Delta m}{\Delta t} g h = 5.0\text{ kg s}^{-1} \times 9.81\text{ m s}^{-2} \times 12.0\text{ m} = 588.6\text{ W}\\n- Rate of gain of kinetic energy:\\n \)P_{\text{KE}} = \frac{1}{2} \frac{\Delta m}{\Delta t} v^2 = \frac{1}{2} \times 5.0\text{ kg s}^{-1} \times (8.0\text{ m s}^{-1})^2 = 160.0\text{ W}\\\n- Total useful power output: \\\n \(P_{\text{useful}} = 588.6\text{ W} + 160.0\text{ W} = 748.6\text{ W}\\n\\n2. Calculate the required electrical power input using efficiency:\n\)\text{Efficiency} = \frac{P_{\text{useful}}}{P_{\text{input}}}\\\n\(0.75 = \frac{748.6\text{ W}}{P_{\text{input}}}\\n\)P_{\text{input}} = \frac{748.6}{0.75} \approx 998\text{ W} \approx 1.0\text{ kW}\)
評分準則
1 mark for the correct option D. Method: Sum the rates of change of potential and kinetic energy to get useful power, then divide by the efficiency.
題目 38 · 選擇題
1 分
Wire X has length \(L\), diameter \(d\), and is made of a metal of Young modulus \(E_X\). When stretched by a force \(F\), it extends by a distance \(x\). Wire Y, made of a metal with Young modulus \(2 E_X\), has length \(2L\) and diameter \(2d\). What is the extension of wire Y when stretched by a force \(2F\)?
A.\(0.25 x\)
B.\(0.5 x\)
C.\(x\)
D.\(2 x\)
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解題
The relationship for the extension \(x\) in terms of Young modulus \(E\), force \(F\), length \(L\), and diameter \(d\) is:\ \(E = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{x / L} = \frac{F L}{\frac{\pi d^2}{4} x} = \frac{4 F L}{\pi d^2 x}\\\n\nRearranging for extension:\n\)x = \frac{4 F L}{\pi d^2 E}\\\\n\\nFor wire X: \(x_X = x = \frac{4 F L}{\pi d^2 E_X}\\\n\nFor wire Y:\n\)x_Y = \frac{4 F_Y L_Y}{\pi d_Y^2 E_Y} = \frac{4 (2F) (2L)}{\pi (2d)^2 (2 E_X)} = \frac{16 F L}{\pi (4 d^2) (2 E_X)} = \frac{16 F L}{8 \pi d^2 E_X} = \frac{2 F L}{\pi d^2 E_X}\\\\n\\nComparing this with \(x_X\):\ \(x_Y = 0.5 \left( \frac{4 F L}{\pi d^2 E_X} \right) = 0.5 x\)
評分準則
1 mark for the correct option B. Method: Set up the formula for extension in terms of the given parameters and substitute the proportional values for Wire Y.
題目 39 · 選擇題
1 分
An ambulance moving at a constant speed along a straight road travels towards a stationary observer. The siren of the ambulance emits a sound of constant frequency \(800\text{ Hz}\). The observer detects a frequency of \(880\text{ Hz}\). The speed of sound in air is \(330\text{ m s}^{-1}\). What is the speed of the ambulance?
A.\(27\text{ m s}^{-1}\)
B.\(30\text{ m s}^{-1}\)
C.\(33\text{ m s}^{-1}\)
D.\(37\text{ m s}^{-1}\)
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解題
Using the Doppler effect formula for a moving source approaching a stationary observer:\ \(f_o = f_s \left( \frac{v}{v - v_s} \right)\\\n\nwhere:\n- \)f_o = 880\text{ Hz}\) (observed frequency)\ - \(f_s = 800\text{ Hz}\) (source frequency)\ - \(v = 330\text{ m s}^{-1}\) (speed of sound)\ - \(v_s\) is the speed of the source.\ \ Substitute the values:\ \(880 = 800 \left( \frac{330}{330 - v_s} \right)\\\n\)1.1 = \frac{330}{330 - v_s}\\\\n\(1.1 (330 - v_s) = 330\\\n\)363 - 1.1 v_s = 330\\\\n\(1.1 v_s = 33\\\n\)v_s = 30\text{ m s}^{-1}\)
評分準則
1 mark for the correct option B. Method: Apply the Doppler formula for a source approaching a stationary observer and solve for the source speed.
題目 40 · 選擇題
1 分
A block of conducting material has a square cross-section of side \(x\) and a length \(L\). The electrical resistance between its two square ends is \(R\). The entire block is melted down and recast into a solid cylinder of length \(2L\). What is the resistance between the flat ends of this cylinder?
A.\(R\)
B.\(2R\)
C.\(4R\)
D.\(8R\)
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解題
1. The original resistance of the block is:\ \(R = \rho \frac{L}{A} = \rho \frac{L}{x^2}\\\n\n2. Let \)V\) be the volume of the block, which remains constant. Thus, the original volume is:\ \(V = x^2 L\\\n\n3. For the recast cylinder of length \)L' = 2L\), its cross-sectional area \(A'\) satisfies:\ \(A' L' = V \implies A' (2L) = x^2 L \implies A' = \frac{x^2}{2}\\\n\n4. The new resistance \)R'\) of the cylinder is:\ \(R' = \rho \frac{L'}{A'} = \rho \frac{2L}{x^2 / 2} = 4 \rho \frac{L}{x^2} = 4R\)
評分準則
1 mark for the correct option C. Method: Use the formula for resistance in terms of resistivity, length, and area, while applying the conservation of volume to find the new area.
卷二 AS Level 結構題
Answer all questions. Show all your working and use appropriate units.
7 題目 · 59.99979999999999 分
題目 1 · Structured
8.5714 分
(a) Define acceleration.
(b) A toy drone rises vertically from rest. The variation of its velocity \(v\) with time \(t\) is described as follows: - From \(t = 0\) to \(t = 4.0\text{ s}\), it accelerates uniformly to a speed of \(6.0\text{ m s}^{-1}\). - From \(t = 4.0\text{ s}\) to \(t = 12.0\text{ s}\), it travels at constant speed. - From \(t = 12.0\text{ s}\) to \(t = 15.0\text{ s}\), it decelerates uniformly to rest.
(i) Calculate the height to which the drone rises during the first \(4.0\text{ s}\).
(ii) Calculate the total height reached by the drone at \(t = 15.0\text{ s}\).
(iii) Determine the magnitude of the deceleration of the drone during the final \(3.0\text{ s}\) of its flight.
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解題
(a) Acceleration is defined as the rate of change of velocity (or change in velocity per unit time).
(b) (i) Height in first 4.0 s is the area under the velocity-time graph from 0 to 4.0 s: \(s = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0 \times 6.0 = 12\text{ m}\).
(ii) Total height is the total area under the velocity-time graph from 0 to 15.0 s (area of a trapezium): \(s = \frac{1}{2} \times (a + b) \times h = \frac{1}{2} \times (8.0 + 15.0) \times 6.0 = 69\text{ m}\).
(iii) Deceleration is the magnitude of the gradient of the graph from 12.0 s to 15.0 s: \(a = \frac{v - u}{t} = \frac{0 - 6.0}{3.0} = -2.0\text{ m s}^{-2}\). Magnitude of deceleration = \(2.0\text{ m s}^{-2}\).
評分準則
Maximum marks: 8 (a) Rate of change of velocity [1] (b)(i) Use of area under graph or \(s = \frac{u+v}{2}t\) [1] Calculation: \(12\text{ m}\) [1] (b)(ii) Area of trapezium formula used or sum of three sections [1] Correct substitution: \(0.5 \times (8.0 + 15.0) \times 6.0\) [1] Calculation: \(69\text{ m}\) [1] (b)(iii) Change in velocity divided by time [1] Calculation: \(2.0\text{ m s}^{-2}\) (ignore negative sign since magnitude is requested) [1]
題目 2 · Structured
8.5714 分
A block of mass \(1.5\text{ kg}\) is sliding on a horizontal frictionless surface at a velocity of \(4.0\text{ m s}^{-1}\). It collides head-on with a second stationary block of mass \(m\). After the collision, the first block rebounds in the opposite direction at a speed of \(0.80\text{ m s}^{-1}\), and the second block moves forward at a speed of \(2.4\text{ m s}^{-1}\).
(a) State the principle of conservation of momentum.
(b) Calculate: (i) the mass \(m\) of the second block, (ii) the loss of total kinetic energy during the collision.
(c) State, with a reason, whether this collision is elastic or inelastic.
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解題
(a) The principle of conservation of momentum states that the total momentum of a system remains constant, provided no external forces act on it.
(b) (i) Let the direction of initial velocity be positive. \(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\) \(1.5 \times 4.0 + 0 = 1.5 \times (-0.80) + m \times 2.4\) \(6.0 = -1.2 + 2.4 m\) \(7.2 = 2.4 m \implies m = 3.0\text{ kg}\).
(c) The collision is inelastic because total kinetic energy is not conserved (kinetic energy is lost/converted to thermal energy).
評分準則
Maximum marks: 8 (a) Total momentum remains constant [1] in an isolated system / when there is no external force [1] (b)(i) Correct momentum conservation equation with negative sign for rebound [1] Substitution: \(6.0 = -1.2 + 2.4m\) [1] Calculation: \(m = 3.0\text{ kg}\) [1] (b)(ii) Calculation of initial KE (\(12\text{ J}\)) and final KE (\(9.12\text{ J}\)) [1] Calculation of KE loss: \(2.9\text{ J}\) (or \(2.88\text{ J}\)) [1] (c) Inelastic AND states that kinetic energy is not conserved [1]
題目 3 · Structured
8.5714 分
(a) Explain the origin of the upthrust acting on an object submerged in a fluid.
(b) A solid metal cylinder of cross-sectional area \(1.5 \times 10^{-3}\text{ m}^2\) and height \(0.20\text{ m}\) is suspended from a spring balance. The cylinder is completely submerged in a container of water of density \(1.0 \times 10^3\text{ kg m}^{-3}\). The bottom of the cylinder is at a depth of \(0.35\text{ m}\) below the surface of the water.
(i) Calculate the difference between the hydrostatic pressure on the bottom face of the cylinder and the hydrostatic pressure on the top face of the cylinder.
(ii) Show that the upthrust acting on the cylinder is \(2.9\text{ N}\).
(iii) The reading on the spring balance when the cylinder is submerged is \(12.5\text{ N}\). Calculate the density of the metal.
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解題
(a) Pressure in a fluid increases with depth because of the weight of fluid above (\(p = \rho g h\)). This means the pressure on the bottom surface of a submerged object is greater than the pressure on its top surface. Since force is pressure multiplied by area, this pressure difference produces an upward force (upthrust) on the object.
(b) (i) The difference in depth is equal to the height of the cylinder, \(h = 0.20\text{ m}\). Pressure difference \(\Delta p = \rho g \Delta h = (1.0 \times 10^3) \times 9.81 \times 0.20 = 1962\text{ Pa} \approx 1.96 \times 10^3\text{ Pa}\) (or \(1.96\text{ kPa}\) or \(2.0\text{ kPa}\)).
(ii) Upthrust \(U = \Delta p \times A = 1962 \times 1.5 \times 10^{-3} = 2.94\text{ N} \approx 2.9\text{ N}\).
(iii) For equilibrium: \(T + U = W\), where \(T\) is tension (spring balance reading) and \(W\) is weight. \(12.5 + 2.94 = W \implies W = 15.44\text{ N}\). Mass of the cylinder: \(M = \frac{W}{g} = \frac{15.44}{9.81} = 1.574\text{ kg}\). Volume of the cylinder: \(V = A \times h = 1.5 \times 10^{-3} \times 0.20 = 3.0 \times 10^{-4}\text{ m}^3\). Density of the metal: \(\rho_m = \frac{M}{V} = \frac{1.574}{3.0 \times 10^{-4}} \approx 5.25 \times 10^3\text{ kg m}^{-3}\) (or \(5.3 \times 10^3\text{ kg m}^{-3}\)).
評分準則
Maximum marks: 8 (a) Pressure increases with depth [1] and results in an upward force because force on the bottom is greater than force on the top [1] (b)(i) Use of \(\Delta p = \rho g h\) [1] Calculation: \(1.96 \times 10^3\text{ Pa}\) (or \(2.0 \times 10^3\text{ Pa}\)) [1] (b)(ii) Use of \(U = \Delta p \times A\) or \(U = \rho g V\) [1] Correct substitution leading to \(2.9\text{ N}\) (minimum 2 sig figs in working) [1] (b)(iii) Calculation of weight: \(W = 12.5 + 2.94 = 15.44\text{ N}\) [1] Calculation of density: \(\rho_m = 5.2 \times 10^3\text{ kg m}^{-3}\) to \(5.3 \times 10^3\text{ kg m}^{-3}\) [1]
題目 4 · Structured
8.5714 分
A block of mass \(0.50\text{ kg}\) is projected up a rough ramp from the bottom with an initial speed of \(6.0\text{ m s}^{-1}\). The ramp is angled at \(30^\circ\) to the horizontal. The block travels a distance of \(2.4\text{ m}\) along the ramp before coming to rest.
(a) Calculate: (i) the initial kinetic energy of the block, (ii) the gain in gravitational potential energy of the block when it comes to rest.
(b) (i) Calculate the work done against the resistive force of friction as the block travels up the ramp. (ii) Determine the average frictional force acting on the block.
(c) Describe the energy changes that occur after the block comes to rest, assuming it slides back down the ramp.
(ii) Height risen \(h = d \sin(30^\circ) = 2.4 \sin(30^\circ) = 1.2\text{ m}\). Gain in gravitational potential energy: \(E_p = m g h = 0.50 \times 9.81 \times 1.2 = 5.89\text{ J} \approx 5.9\text{ J}\).
(ii) Work done against friction is \(W_f = F_f \times d\). \(3.11 = F_f \times 2.4 \implies F_f = \frac{3.11}{2.4} \approx 1.3\text{ N}\).
(c) As the block slides back down, gravitational potential energy is converted into kinetic energy and thermal energy (work done against friction increases the internal energy of the block and ramp).
評分準則
Maximum marks: 8 (a)(i) Use of \(E_k = \frac{1}{2} m v^2\) [1] Calculation: \(9.0\text{ J}\) [1] (a)(ii) Use of \(h = d \sin(30^\circ)\) [1] Calculation: \(5.9\text{ J}\) (or \(5.89\text{ J}\)) [1] (b)(i) subtraction: \(9.0 - 5.89\) [1] Calculation: \(3.1\text{ J}\) (or \(3.11\text{ J}\)) [1] (b)(ii) Use of \(W_f = F_f \times d\) to find \(F_f = 1.3\text{ N}\) [1] (c) Gravitational potential energy converted to kinetic energy and thermal energy [1]
題目 5 · Structured
8.5714 分
(a) Describe what is meant by a progressive transverse wave.
(b) A progressive transverse wave travels along a horizontal string. The wave has a frequency of \(25\text{ Hz}\). The distance between a peak and the adjacent trough is \(0.40\text{ m}\) along the horizontal direction of propagation, and the vertical distance from peak to trough is \(8.0\text{ mm}\).
Calculate: (i) the amplitude of the wave, (ii) the wavelength of the wave, (iii) the speed of the wave.
(c) Two points, P and Q, on the string are separated by a horizontal distance of \(0.15\text{ m}\). Calculate the phase difference, in degrees, between the oscillations of P and Q.
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解題
(a) A progressive wave is a disturbance that transfers energy from one point to another without transferring matter. A transverse wave is one where the particles oscillate perpendicular to the direction of energy transfer.
(b) (i) Amplitude is half the vertical distance from peak to trough: \(A = \frac{8.0\text{ mm}}{2} = 4.0\text{ mm}\) (or \(4.0 \times 10^{-3}\text{ m}\)).
(ii) The horizontal distance from peak to adjacent trough is half of one wavelength: \(\frac{\lambda}{2} = 0.40\text{ m} \implies \lambda = 0.80\text{ m}\).
(iii) Speed \(v = f \lambda = 25 \times 0.80 = 20\text{ m s}^{-1}\).
Maximum marks: 8 (a) Transfer of energy without transfer of matter [1] Oscillations are perpendicular to direction of wave travel / energy transfer [1] (b)(i) \(4.0\text{ mm}\) (or \(4.0 \times 10^{-3}\text{ m}\)) [1] (b)(ii) \(0.80\text{ m}\) [1] (b)(iii) Use of \(v = f \lambda\) [1] Calculation: \(20\text{ m s}^{-1}\) [1] (c) Use of \(\text{phase difference} = \frac{x}{\lambda} \times 360^\circ\) [1] Calculation: \(67.5^\circ\) or \(68^\circ\) (accept \(1.18\text{ rad}\)) [1]
題目 6 · Structured
8.5714 分
(a) Define: (i) resistance, (ii) resistivity.
(b) A uniform wire of resistance \(4.8\ \Omega\) has a length of \(1.5\text{ m}\) and a cross-sectional area of \(2.4 \times 10^{-7}\text{ m}^2\).
(i) Calculate the resistivity of the material of the wire.
(ii) The wire is stretched uniformly so that its length increases to \(3.0\text{ m}\), while its volume remains constant. Calculate the new resistance of the wire.
(iii) The original wire and the stretched wire are connected in turn across the same potential difference. Explain, without calculation, how the drift speed of the free electrons changes after the wire is stretched.
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解題
(a) (i) Resistance is defined as the ratio of potential difference across a conductor to the current through it (\(R = V/I\)).
(ii) Resistivity is a property of the material defined by \(\rho = \frac{R A}{L}\), where \(R\) is resistance, \(A\) is cross-sectional area, and \(L\) is length.
(ii) Volume \(V = A \times L\) is constant. Since \(L\) is doubled (from \(1.5\text{ m}\) to \(3.0\text{ m}\)), the cross-sectional area \(A\) must be halved to keep volume constant. Resistance \(R = \frac{\rho L}{A}\). Since \(L \rightarrow 2L\) and \(A \rightarrow A/2\), the new resistance is: \(R_{new} = \frac{\rho (2L)}{A/2} = 4 \frac{\rho L}{A} = 4 \times R = 4 \times 4.8 = 19.2\ \Omega \approx 19\ \Omega\).
(iii) Current is given by \(I = n A v e\) and also by \(I = V/R\). Therefore, \(v = \frac{V}{R n A e}\). When stretched, \(R\) increases by a factor of 4 and \(A\) is halved, so the product \(R \times A\) increases by a factor of 2. Since \(V\), \(n\), and \(e\) are constant, the drift speed \(v\) must decrease by a factor of 2. Alternatively: The electric field strength in the wire is \(E = V/L\). Since length \(L\) is doubled for the same \(V\), the electric field strength is halved, which halves the force on each electron and thus halves the drift speed.
評分準則
Maximum marks: 8 (a)(i) Ratio of potential difference to current [1] (a)(ii) \(\rho = R A / L\) with symbols defined [1] (b)(i) Correct formula used [1] Calculation: \(7.7 \times 10^{-7}\ \Omega\text{ m}\) (accept \(7.68 \times 10^{-7}\ \Omega\text{ m}\)) [1] (b)(ii) Statement or use of \(A\) halving when \(L\) doubles [1] Calculation: \(19\ \Omega\) or \(19.2\ \Omega\) [1] (b)(iii) Reference to \(I = n A v e\) or \(E = V/L\) [1] Explanation showing drift speed decreases (by factor of 2) [1]
題目 7 · Structured
8.5714 分
(a) Hadrons are subatomic particles made of quarks. State the names of the two main classes of hadrons and describe the quark composition of each class.
(b) A free neutron is unstable and decays according to the equation: \[\text{n} \rightarrow \text{p} + \text{e}^- + \overline{\nu}_\text{e}\]
(i) State the name of the fundamental interaction responsible for this decay.
(ii) Describe the change in quark flavour that occurs during this decay.
(iii) By considering the charges of the quarks, show that charge is conserved in this decay.
(c) State the lepton number of each particle in the decay to show that lepton number is conserved.
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解題
(a) The two main classes of hadrons are: - Baryons, which consist of three quarks (or three antiquarks for antibaryons). - Mesons, which consist of a quark and an antiquark.
(b) (i) The fundamental interaction responsible is the weak interaction (or weak nuclear force).
(ii) In a neutron (quark structure \(\text{udd}\)) decaying to a proton (quark structure \(\text{uud}\)), one down quark (\(\text{d}\)) changes into an up quark (\(\text{u}\)).
(iii) Quark charges: Up quark (\(\text{u}\)) = \(+\frac{2}{3} e\) Down quark (\(\text{d}\)) = \(-\frac{1}{3} e\) Charge of neutron (\(\text{udd}\)) = \(+\frac{2}{3} e - \frac{1}{3} e - \frac{1}{3} e = 0\). Charge of proton (\(\text{uud}\)) = \(+\frac{2}{3} e + \frac{2}{3} e - \frac{1}{3} e = +1 e\). Total charge on left side = 0. Total charge on right side = \(\text{proton charge} + \text{electron charge} + \text{antineutrino charge} = +1e - 1e + 0 = 0\). Since both sides equal 0, charge is conserved.
(c) Lepton numbers of each particle: - Neutron (\(\text{n}\)): 0 (not a lepton) - Proton (\(\text{p}\)): 0 (not a lepton) - Electron (\(\text{e}^-\)): \(+1\) - Electron antineutrino (\(\overline{\nu}_\text{e}\)): \(-1\) Total lepton number before decay = 0. Total lepton number after decay = \(0 + (+1) + (-1) = 0\). Since the total lepton number is 0 before and after, lepton number is conserved.
評分準則
Maximum marks: 8 (a) Baryons and Mesons [1] Baryons have three quarks AND mesons have a quark and an antiquark [1] (b)(i) Weak interaction / weak nuclear force [1] (b)(ii) Down quark (\(d\)) to up quark (\(u\)) [1] (b)(iii) Show neutron charge is 0 using quark charges [1] Show proton charge is +1 and state total final charge is 0, concluding conservation [1] (c) Correctly identify neutron and proton lepton number as 0 [1] Correctly identify electron as +1 and antineutrino as -1, showing sum is 0 [1]
Paper 3 Advanced Practical Skills
Answer both questions. Record all your observations as soon as they are made.
2 題目 · 40 分
題目 1 · Practical
20 分
### Question 1: Quantitative Task
In this experiment, you will investigate how the height of a pivoted, spring-supported metre rule depends on the position of a load.
#### Apparatus and Materials - Metre rule (used as the horizontal pivoted beam) - Vertical metre rule (used to measure height) - Stand, boss and clamp to hold the vertical metre rule - Stand, boss and clamp to suspend a spring - Tension spring with a stiffness in the range of \(10\text{ to }25\text{ N m}^{-1}\) - Thread loop attached at the \(0.0\text{ cm}\) mark of the horizontal rule to act as a pivot - Pin or nail held in a clamp to act as the pivot axle - Heavy mass \(m = 0.200\text{ kg}\) (e.g. a mass hanger with slotted masses) - Set square
#### Fig. 1.1
``` [Stand] | | [Spring] |==== S ==== | | | (Pivot) [Rule] | | | ======O==============|====|====|====[ 90.0 cm ] | | | | | [Stand] | v v | [Mass m] v [Vertical Rule] ---------------------------------------------------- [Bench] ```
#### Procedure
1. Set up the apparatus as shown in Fig. 1.1, ensuring that: - The pivot at the \(0.0\text{ cm}\) mark is secure and allows the rule to rotate freely in the vertical plane. - The spring is suspended vertically from its clamp and is attached to the \(90.0\text{ cm}\) mark of the horizontal rule. - The horizontal rule is approximately horizontal when there is no mass on it. Adjust the height of the spring's support if necessary. - The vertical metre rule is placed next to the \(90.0\text{ cm}\) mark of the horizontal rule. Use the set square to ensure this measuring rule is vertical.
2. Record the height \(h_0\) of the top edge of the horizontal rule at the \(90.0\text{ cm}\) mark above the laboratory bench when there is no mass on the rule.
3. Place the mass hanger of mass \(m = 0.200\text{ kg}\) at a distance \(d = 10.0\text{ cm}\) from the pivot (i.e., at the \(10.0\text{ cm}\) mark of the rule).
4. Record the new height \(h\) of the \(90.0\text{ cm}\) mark.
5. Calculate the depression \(y\) of this mark, where: \[ y = h_0 - h \]
6. Vary \(d\) in the range \(10.0\text{ cm} \le d \le 80.0\text{ cm}\) and repeat steps 3 to 5 to obtain six sets of values for \(d\), \(h\), and \(y\).
7. Tabulate all your observations. Include columns for \(d\), \(h\), and \(y\), ensuring correct units for each heading.
8. Plot a graph of \(y\) on the y-axis against \(d\) on the x-axis. Draw the straight line of best fit.
9. Determine the gradient and y-intercept of this line.
10. The quantities \(y\) and \(d\) are related by the equation: \[ y = a d + b \] where \(a\) and \(b\) are constants. State the values of \(a\) and \(b\). If you plotted \(y\) and \(d\) in metres, determine the units of \(a\) and \(b\).
11. Use your value of \(a\) to determine the spring constant \(k\) of the spring, given that: \[ a = \frac{m g}{k L} \] where \(m = 0.200\text{ kg}\), \(g = 9.81\text{ m s}^{-2}\), and \(L = 0.900\text{ m}\).
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解題
### Worked Solution
#### 1. Sample Experimental Data Let the vertical height of the horizontal rule with no mass added be \(h_0 = 40.0\text{ cm} = 0.400\text{ m}\).
When a mass \(m = 0.200\text{ kg}\) is placed at various distances \(d\) from the pivot, the raw height \(h\) is measured and the depression \(y\) is calculated as follows:
#### 2. Graph Plotting - The independent variable \(d\) is plotted on the horizontal axis (x-axis). - The dependent variable \(y\) is plotted on the vertical axis (y-axis). - Let the scale be: - x-axis: \(1\text{ large grid square } (2\text{ cm}) = 0.1\text{ m}\) - y-axis: \(1\text{ large grid square } (2\text{ cm}) = 0.01\text{ m}\) - The plotted points lie on a clear, straight line passing close to the origin.
#### 3. Gradient and Intercept Calculation Using two distant points on the line of best fit: - Point 1: \((0.150\text{ m}, 0.016\text{ m})\) - Point 2: \((0.750\text{ m}, 0.082\text{ m})\)
Since \(y\) and \(d\) are both in metres, the gradient \(a\) is dimensionless: \[ a = 0.110 \]
Using the point \((0.150, 0.016)\) to find the y-intercept \(b\): \[ b = y - a d = 0.016 - (0.110 \times 0.150) = 0.016 - 0.0165 = -0.0005\text{ m} \approx 0.000\text{ m} \]
#### 4. Calculating Spring Constant \(k\) Using the formula: \[ a = \frac{m g}{k L} \implies k = \frac{m g}{a L} \]
Substitute \(m = 0.200\text{ kg}\), \(g = 9.81\text{ m s}^{-2}\), \(L = 0.900\text{ m}\), and \(a = 0.110\): \[ k = \frac{0.200 \times 9.81}{0.110 \times 0.900} = \frac{1.962}{0.099} = 19.8\text{ N m}^{-1} \]
(This matches the expected spring constant value of approximately \(20\text{ N m}^{-1}\)).
評分準則
### Marking Scheme
#### (a) Table of results (6 marks) - **[1]** Table includes headings with correct units: \(d / \text{m}\) (or \(\text{cm}\)), \(h / \text{m}\), and \(y / \text{m}\). - **[1]** All raw measurements of height \(h\) are recorded to the nearest millimetre (consistent to 3 decimal places if in metres, or 1 decimal place if in cm). - **[1]** Correct calculation of \(y = h_0 - h\) with consistent decimal places. - **[1]** At least six sets of readings recorded, spanning a range of \(d\) of at least \(0.500\text{ m}\) (e.g., from \(0.100\text{ m}\) to \(0.700\text{ m}\) or more). - **[1]** Raw values show the correct trend (as \(d\) increases, \(h\) decreases and \(y\) increases). - **[1]** Quality of data: All points lie within \(\pm 2\text{ mm}\) of the line of best fit vertically.
#### (b) Graph (4 marks) - **[1]** Axes: properly labeled with quantities and units. Scales chosen so that the plotted points occupy more than half of the grid in both directions. - **[1]** Plotting: Points are plotted accurately to within half a small grid square. No large blobs or thick pencil marks. - **[1]** Line of best fit: A single straight line drawn with a ruler, representing a fair compromise of the plotted data, with an even distribution of points on either side of the line. - **[1]** Precision of line: Line thickness is thin and continuous, with no feathered lines.
#### (c) Gradient and Intercept (2 marks) - **[1]** Gradient: Determined using a large triangle where the hypotenuse is longer than half the length of the drawn line. Coordinates of points on the line read off correctly to within half a small square. - **[1]** Intercept: Determined from direct read-off if the x-axis starts at \(d = 0\), or calculated using a point on the line in \(y = a d + b\).
#### (d) Constants \(a\) and \(b\) (4 marks) - **[1]** Value of \(a\) is equated to the gradient. - **[1]** Value of \(b\) is equated to the y-intercept. - **[1]** Unit of \(a\) is correctly stated as dimensionless (or \(\text{m m}^{-1}\)) and unit of \(b\) is correctly stated as \(\text{m}\) (or \(\text{cm}\)). - **[1]** Both \(a\) and \(b\) are stated to an appropriate number of significant figures (usually 2 or 3).
#### (e) Spring Constant \(k\) (4 marks) - **[1]** Correct algebraic rearrangement of the expression to make \(k\) the subject: \(k = \frac{m g}{a L}\). - **[1]** Correct substitution of the student's value of \(a\), alongside the given parameters \(m = 0.200\text{ kg}\), \(g = 9.81\text{ m s}^{-2}\), and \(L = 0.900\text{ m}\). - **[1]** Spring constant \(k\) calculated correctly with the correct unit: \(\text{N m}^{-1}\). - **[1]** Value of \(k\) is written to 2 or 3 significant figures, matching the precision of the experimental data.
題目 2 · Practical
20 分
### Question 2: Investigation and Evaluation
In this experiment, you will investigate how the sliding distance of a plastic cup across a table depends on the release position of a rolling steel ball that collides with it.
#### Apparatus and Materials - Metre rule (used as an inclined runway) - Support block (approx. \(5\text{ cm}\) to \(10\text{ cm}\) tall) to elevate one end of the runway - Small paper or plastic cup (approx. \(150\text{ mL}\)) - A second metre rule (used to measure distances) - Smooth heavy steel ball (or large marble) of diameter approx. \(2.0\text{ cm}\) - Set square
#### Fig. 2.1
``` [Runway Rule] \ \ <- [Steel ball released from distance x] \ \ =======\=======================\================== [Bench] \ | | \====================v | [Cup] v [Sliding distance s] ```
#### Procedure
1. Set up the runway by resting one end of a metre rule on the support block and the other end on the horizontal table bench, as shown in Fig. 2.1.
2. Place the cup on the bench with its opening facing towards the bottom of the runway. The cup's center should be aligned with the path of a ball rolling down the runway, at a distance of \(5.0\text{ cm}\) from the end of the runway rule.
3. Mark a release point on the runway at a distance \(x_1 = 30.0\text{ cm}\) from the bottom end of the runway rule.
4. Hold the steel ball at this release point, and release it from rest.
5. The ball will roll down, enter the cup, and the cup will slide along the table bench and come to rest.
6. Measure the distance \(s_1\) that the cup has slid. (This is the distance from the initial position of the cup to its final stopping position, measured along the line of travel).
7. Perform at least two trials for \(x_1 = 30.0\text{ cm}\) and determine the average sliding distance \(s_1\).
8. Estimate the percentage uncertainty in your measurement of \(s_1\). Show your working.
9. Repeat the procedure using a second release distance \(x_2 = 60.0\text{ cm}\). Record your raw trials and calculate the average sliding distance \(s_2\).
10. It is suggested that the sliding distance \(s\) is directly proportional to the release distance \(x\), so that: \[ s = k x \] where \(k\) is a constant. Calculate the values of the constant \(k\) for both sets of measurements, writing them as \(k_1\) and \(k_2\).
11. State whether your results support the suggestion. Explain your reasoning clearly, using a quantitative comparison of the percentage difference between \(k_1\) and \(k_2\) with a sensible experimental uncertainty limit (e.g., \(10\%\)).
12. Identify four limitations of the experiment and suggest four corresponding improvements to reduce these sources of error. Present your answers in a table.
#### 2. Percentage Uncertainty in \(s_1\) - Let the absolute uncertainty in the stopping position be \(\Delta s = 0.5\text{ cm}\) (due to the difficulty in keeping the cup aligned and judging the exact friction coefficient of the table). - Percentage uncertainty: \[ \%\text{ Uncertainty} = \frac{\Delta s_1}{s_1} \times 100\% = \frac{0.5}{18.4} \times 100\% = 2.72\% \approx 2.7\% \]
#### 4. Calculation of constants \(k_1\) and \(k_2\) Using \(s = k x\): - For \(x_1 = 30.0\text{ cm} = 0.300\text{ m}\): \[ k_1 = \frac{s_1}{x_1} = \frac{18.4}{30.0} = 0.613 \] - For \(x_2 = 60.0\text{ cm} = 0.600\text{ m}\): \[ k_2 = \frac{s_2}{x_2} = \frac{35.8}{60.0} = 0.597 \]
#### 5. Quantitative Analysis and Conclusion - Percentage difference between \(k_1\) and \(k_2\): \[ \%\text{ Difference} = \frac{|k_1 - k_2|}{k_1} \times 100\% = \frac{|0.613 - 0.597|}{0.613} \times 100\% = \frac{0.016}{0.613} \times 100\% = 2.61\% \approx 2.6\% \] - Comparison: The percentage difference of \(2.6\%\) is significantly smaller than the typical experimental limit of \(10\%\). - Conclusion: The results strongly support the suggestion that the sliding distance is directly proportional to the release distance.
評分準則
### Marking Scheme
#### (a) First set of measurements (2 marks) - **[1]** Raw trial values of \(s_{1a}\) and \(s_{1b}\) are recorded to the nearest millimetre with correct units (\(\text{cm}\) or \(\text{m}\)). - **[1]** Average value \(s_1\) is calculated correctly from these trials.
#### (b) Percentage Uncertainty (2 marks) - **[1]** Sensible absolute uncertainty chosen: \(\Delta s\) should be in the range \(0.5\text{ cm}\) to \(1.0\text{ cm}\) with an explanation (e.g., "due to difficult alignment of the cup"). - **[1]** Correct percentage calculation: \(\frac{\Delta s_1}{s_1} \times 100\%\) (e.g., around \(2.7\%\) to \(5.4\%\)).
#### (c) Second set of measurements (2 marks) - **[1]** Raw trial values of \(s_{2a}\) and \(s_{2b}\) recorded to the nearest millimetre. - **[1]** Average value \(s_2\) is calculated correctly.
#### (d) Precision and Calculations (2 marks) - **[1]** All raw values of \(x_1\), \(x_2\), \(s_1\), and \(s_2\) are recorded consistently to the nearest millimetre. - **[1]** Constants \(k_1\) and \(k_2\) calculated correctly with no units (if both distances are in the same units) and stated to 2 or 3 significant figures.
#### (e) Analysis and Conclusion (4 marks) - **[1]** Percentage difference between \(k_1\) and \(k_2\) calculated correctly. - **[1]** Correct comparison stated (e.g., "The percentage difference is \(2.6\%\)"). - **[1]** An experimental threshold limit is clearly stated (e.g., "I will assume a threshold of \(10\%\)"). - **[1]** Clear concluding statement matching the math: "Since the percentage difference of \(2.6\%\) is less than the \(10\%\) limit, the hypothesis that \(s \propto x\) is supported."
#### (f) Evaluation (8 marks total) *Award 1 mark for each limitation (up to 4) and 1 mark for each corresponding improvement (up to 4):*
1. **[1] Limitation:** Difficult to release the ball consistently from rest without giving it an initial push or sideways spin. **[1] Improvement:** Use an electromagnetic clamp, a mechanical spring gate, or a thin plastic ruler to act as a trapdoor release mechanism.
2. **[1] Limitation:** The cup slides at an angle or rotates slightly, making the distance measurement curved rather than straight. **[1] Improvement:** Use two parallel guide tracks (such as two long rules taped to the bench) to keep the cup sliding in a straight line.
3. **[1] Limitation:** Difficult to measure the final position of the cup accurately because the cup decelerates slowly and the stopping point is hard to judge visually. **[1] Improvement:** Position a video camera (high frame-rate smartphone camera) over the bench with a ruler placed alongside the path, and play back frame-by-frame to see where motion ceases.
4. **[1] Limitation:** Two sets of readings are not enough to confirm a general linear relationship. **[1] Improvement:** Take multiple readings of \(s\) for several different release distances \(x\) (at least six different positions) and plot a graph of \(s\) against \(x\).
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